Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation? 
Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation?

Rank is defined to be $\dim(\text{im}(T))$ right? Every output of $T$ is a vector belonging to $\Bbb{R^2}$ and $\Bbb{R^2}$ is the image of $T$ so I assumed that $\text{rank}(T) = 2$. Turns out it's actually $1$, which I don't seem to understand... Every output of $T$ can be expressed as $c(1,2)$ but this is also just a tuple in $\Bbb{R^2}$ so how can the rank be $1$?
The nullity is defined to be the set of vectors for which $T(x) = 0$ so I have that $$(x+y,2(x+y)) = 0 \Longrightarrow x+y=0, 2x+2y=0 \Longrightarrow (x,y) = (0,0)$$
so the set of vectors for which $T(x) = 0$ is just $\{(0,0)\}$ so $\ker(T) = 1$?
 A: When calculating the rank of a linear transformation you're interested in the dimension of its image. Then nullity is the dimension of the kernel, that is all the vectors that are mapped to the zero vector by the transformation. The rank plus the nullity will be the dimension of the domain.
Lets consider a simple example first, the zero transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y)=(0,0)$. Here all the vectors are transformed into the zero vector, which is a zero dimensional space. Another simple example is the identity $T(x,y)=(x,y)$ and the image is clearly the entire plane, so it has rank $2$ and nullity $0$.
Now lets look at one that is almost your example with $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y)=(x+y,x+y)$. If we set $a=x+y$ we see that these points are all of the form $(a,a)$, which is diagonal in the plane. This line is one dimensional and so the image of the transformation is one dimensional, giving it rank $1$. Since we know that the dimension of the image and kernel must sum to $2$ this also tells us that the nullity is $1$. See if you can find the line through the origin that is the kernel of this transformation.
Finally your transformation is $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y)=(x+y,2(x+y))$. Again setting $a= x+y$ we now have points of the form $(a,2a)$ which is again a line and so the image has rank $1$ and we calculate that the nullity is $1$ as well. Using some trial and error see if you can find the kernel, which is another line in the plane. You can test them by ensuring $T(x,y)=0$.
Now in all these cases you could set up the matrix and do row reduction as well to find the dimension. The number of non-zero pivots is the dimension of the image. All matrices has four fundamental subspaces, the two we discussed being the image and kernel, and the same two of its transpose.
A: Hint:
The matrix for this linear map is
$$
\begin{pmatrix}
1&1\\
2&2
\end{pmatrix}
$$
which is singular.

Notes 1. $x+y=0$ and $2x+2y=0$ implies that $x=-y$, not $x=y=0$. So the kernel of your map is $\operatorname{span}\{(1,-1)\}$ and thus the nullity is $1$. Now you can apply the rank-nullity theorem to find the rank or directly tells it from the matrix.

Notes 2.

*

*The rank of a linear transformation is the dimension of the image. That's correct.

*Every output of $T$ is a vector in $\mathbb{R}^2$ does not mean that the image of $T$ is $\mathbb{R}^2$.

A: Note that $$T(x,y) = (x+y) (1,2),$$so the range of $T$ is spanned by $(1,2)$. But a non-zero vector by itself is linearly independent, so $\{(1,2)\}$ is a basis for the range, and thus ${\rm rank}(T) = 1$. By the rank-nullity theorem, $\dim \ker T = 1$ as well, and so any non-zero vector in the kernel of $T$ forms a basis for it, by the same argument above: clearly $(1,-1)$ fits the bill.
